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\title[On Pro-$p$ Cappitt Groups with finite exponent] {On Pro-$p$ Cappitt Groups with finite exponent}

\author{\firstname{Anderson} \lastname{Porto}\CDRorcid{0000-0002-8800-0827}}
\address{Instituto de Ciência e Tecnologia - ICT, Universidade Federal dos Vales do Jequitinhonha e Mucuri, Diamantina - MG, 39100-000 Brazil}
\address{Departamento de Matem\'atica, Universidade de Bras\'ilia, Brasilia-DF, 70910-900 Brazil}
\email{ander.porto@ict.ufvjm.edu.br}

\author{\firstname{Igor} \lastname{Lima}\IsCorresp}
\address[1]{Instituto de Ciência e Tecnologia - ICT, Universidade Federal dos Vales do Jequitinhonha e Mucuri, Diamantina - MG, 39100-000 Brazil}
\address[2]{Departamento de Matem\'atica, Universidade de Bras\'ilia, Brasilia-DF, 70910-900 Brazil}
\email{igor.matematico@gmail.com}

\subjclass{20E34, 20E18}

\keywords{Generalized Dedekind groups, pro-$p$ Cappitt groups, torsion groups.}

\thanks{The first author was partially supported by FAPDF, Brazil. The second author was partially supported by DPI/UnB, FEMAT Proc. 054/2022, FAPDF, Brazil.}

\dedicatory{To Pavel Zalesskii on his 60th birthday}

\begin{abstract}
A pro-$p$ Cappitt group is a pro-$p$ group $G$ such that $\tilde{S}(G) = \overline{ \langle L \leqslant_c G \:|\, L \ntriangleleft G \rangle}$ is a proper subgroup (i.e. $\tilde{S}(G) \neq G$). In this paper we prove that non-abelian pro-$p$ Cappitt groups whose torsion subgroup is closed and it has finite exponent. This result is a natural continuation of main result of the first author~\cite{Por19}. We also prove that in a pro-$p$ Cappitt group its subgroup commutator is a procyclic central subgroup. Finally we show that pro-$2$ Cappitt groups of exponent $4$ are pro-$2$ Dedekind groups. These results are pro-$p$ versions of the generalized Dedekind groups studied by Cappitt (see Theorem~1 and Lemma~7 in~\cite{Cap71}).
\end{abstract}

\dateposted{2024-05-02}
\begin{document}
\maketitle

\section{Introduction}

Profinite groups are totally disconnected compact Hausdorff topological groups. Such groups can be seen as projective limits of finite groups (see Ribes--Zalesskii~\cite{RZ10} or Wilson~\cite{wilson98}).

We recall that a totally disconnected compact Hausdorff group $G$ is called a profinite torsion group (or periodic) if all of its elements are of finite order.

One motivation for studying these groups is the following open question due Hewitt--Ross (see Open Question~4.8.5b in~\cite{RZ10}): Is a profinite torsion group necessarily of finite exponent?

In celebrated paper~\cite{Zelm92}, Zel$'$manov proves that a finitely generated pro-$p$ torsion group is finite. Therefore the question above is open for infinitely generated torsion groups.

In~\cite{Herf80} and~\cite{Herf82}, Herfort proves that a profinite group $G$ whose order is divisible by infinitely many different primes has a procyclic subgroup with the same property. Otherwise if $G$ is a profinite torsion group then the order of $G$ is divisible by only finitely many distinct primes.


As main result in this paper we prove that a infinitely generated non-abelian pro-$p$ Cappitt group has finite exponent. This result is the profinite version of second part of Theorem~1 in~\cite{Cap71}. In~\cite{Por19} the first author of this paper proves the first part of the our main result (see Theorem~\ref{teoA}). Also we obtained other related results of independent interest (see Theorem~\ref{teoB}).


\section{Preliminaries}

The standard notation is in accordance with~\cite{Por19}, \cite{RZ10} and~\cite{wilson98}.

We recall that an abstract group (finite or infinite) is called a \emph{Dedekind group} if every subgroup is normal. As in~\cite{Cap71}, for any group $G$, we define $S(G)$ to be the subgroup of $G$ generated by all the subgroups which are not normal in $G$, i.e. $S(G) = \langle N \,|\, N \leq G \text{ and } N \not\trianglelefteq G \rangle$. Note that $G$ is a Dedekind group if and only if $S(G)=\{1\}$. Also $S(G)$ is a characteristic subgroup of $G$ and the quotient group of $G$ by $S(G)$ is a Dedekind group. A group is to be a \emph{Cappitt group} if it satisfies $S(G)\neq G$, these groups are generalizations of Dedekind groups (see for example Kappe and Reboli~\cite{kappe} or Cappitt~\cite{Cap71}).


In the profinite version, as in~\cite{Por19} and~\cite{PB19}, a \emph{profinite Dedekind group} will be a profinite group in which every closed subgroup is normal. A profinite non-abelian Dedekind group will be called a \emph{profinite Hamiltonian group}.

\begin{rema}\label{remark prociclica}
If each procyclic subgroup of a profinite group $G$ is normal, then G is a profinite Dedekind group (see Remark~1, p.~90-91 in~\cite{PB19}).
\end{rema}

A. Porto and V. Bessa~\cite{PB19} classified these groups as follow

\begin{theo}
A profinite group $G$ is Dedekind if, and only if, $G$ is abelian or there exists a finite set of odd primes $J$ and a natural number $e$ such that
\[
G \cong Q_8 \times \tilde{E} \times \prod_{p \in J} \left(\prod_{i=1}^{e} \left(\prod_{m(i,p)}\,C_{p^i}\right)\right),
\]
where $Q_8$ is the quaternion group of order $8,$ $\tilde{E}$ is an elementary abelian pro-$2$ group and each $m(i,p)$ is a cardinal number. In particular, if $G$ is a profinite Hamiltonian group then $G$ has finite exponent.\label{dedekindBP}
\end{theo}


Define $G\,'$ as the topological closure in $G$ of the abstract commutator subgroup $[G, G] = \langle [g,h]\,|\, g,h \in G \rangle$, where $[g,h]=g^{-1}h^{-1}gh$ is the commutator of the elements $g, h \in G$. The abelianization of group $G$ will be denoted by $G^{ab} = G/G'=G/\overline{[G,G]}$. If $G$ is a profinite group, $G^{ab}$ is an abelian profinite group.


\begin{defi*}
A \emph{profinite Cappitt group} is a profinite group $G$ such that $\tilde{S}(G) = \overline{ \langle L \leqslant_c G \,|\,\, L \ntriangleleft G \rangle}$ is a proper subgroup of $G$.
\end{defi*}

This profinite version was defined firstly in~\cite{Por19}. For finite groups we have that $\tilde{S}(G)=S(G).$

Using Theorem~\ref{dedekindBP}, Porto gave the following characterization of profinite Cappitt groups.

\begin{theo}
Let $G$ be a profinite group satisfying $\{1\} \neq \tilde{S}(G) \neq G.$ Then $G$ can be expressed as direct product $\tilde{H} \times \tilde{K}$, where $\tilde{H}$ is a pro-$p$ Cappitt group for some prime $p$ and $\tilde{K}$ is a profinite Dedekind group that does not contain elements of order $p$. Such $G$ is nilpotent profinite of class at most $2$.\label{mainPorto}
\end{theo}

Denote by $\mathbb{Z}_{p}$ the $p$-adic integers. Consider $|z|$ the order of an element $z$ of a group $G$. For a group $G$ we denote $\tor(G) = \{x \in G\,|\, |x| < \infty \}$ the subset of the elements of torsion of $G$. As in the abstract case (see 16.2.7 in~\cite{Kargapolov}), when $G$ is a nilpotent pro-$p$ group, then $\tor(G)$ is a subgroup of $G$ (not necessarily closed). A main result of this paper is the following

\begin{ltheo} \label{teoA}
Let $G$ be a non-abelian pro-$p$ Cappitt group. Then $G\,'$ is a procyclic central subgroup. Moreover, if $\tor(G) \leqslant_{c} G$ then $G$ has finite exponent.
\end{ltheo}


\begin{rema}\label{matalogo}
The Theorem~\ref{teoA} has immediate proof when $G$ is a pro-$2$ Hamiltonian group since $G\,' = C_2$ is central because $G$ has nilpotent class $2$ (see Corollary~$1$ in~\cite{Por19}), and also $G$ has finite exponent by Theorem~\ref{dedekindBP}. The condition to be non-abelian is necessary in the Theorem~\ref{teoA} since $\mathbb{Z}_{p}$ is an abelian non-periodic group satisfying all hypothesis.
\end{rema}

Now we present some preliminaries results that will are used throughout in the proof of the main theorem.

\begin{lemm}\label{limitesdeS}
Let $G = \varprojlim_{i \in \mathcal{I}} \{G_i, \phi_{ij}\}$ where $\{G_i, \phi_{ij}, \mathcal{I}\}$ is a surjective inverse system of finite groups (with discrete topology). Then
\[
\tilde{S}(G) = \varprojlim_{i \in \mathcal{I}} \{S(G_i), \phi_{ij}\vert_{_{S(G_i)}} \},
\]
where $\{S(G_i), \phi_{ij}\vert_{_{S(G_i)}}, \mathcal{I}\}$ is the associated surjective inverse system.
\end{lemm}

\begin{proof}
See Proposition~$1$ in~\cite{Por19}.
\end{proof}

The following facts are direct consequences of the definition of subgroups $S(G), S(L),\ S(G / J)$ and the Correspondence Theorem. In the case of abstract groups, these can be found in remark on page 312 in~\cite{Cap71} or Lemma~2.2 in~\cite{kappe}.

\begin{rema}\label{quocientecappitt}
If $G$ is an abstract group with $S(G) \neq G$ and $J$ is a normal subgroup of $G$ contained in $S(G)$, then $S(G/J) \neq G/J$, furthermore if $L$ is a subgroup of $G$ containing elements of $G \setminus S(G)$ then $S(L) \neq L$.
\end{rema}

Analogously to the abstract case, similar results can be shown in the category of profinite groups, as follows.

\begin{rema}\label{quocientecappittprofinite}
If $G$ is a profinite Cappitt group (i.e. $\tilde{S}(G) \neq G$) and $J$ is a closed normal subgroup of $G$ contained in $\tilde{S}(G)$, then $\tilde{S}(G/J) \neq G/J$, furthermore if $L$ is a closed subgroup of $G$ containing elements of $G \setminus \tilde{S}(G)$ then $\tilde{S}(L) \neq L$.
\end{rema}



The following result is similar to Lemma~$1$ in~\cite{Por19}, we will repeat the same argument for the convenience of the reader.

\begin{lemm}
Let $G$ be as in the Lemma~\ref{limitesdeS}. If $G \neq \tilde{S}(G)$ then $G_i \neq S(G_i),\,\, \forall i \in \mathcal{I}$.
\label{commutador2}
\end{lemm}

\begin{proof}
Suppose that there is a $G_i$ such that $\tilde{S}(G_i)=S(G_i) = G_i$. Denote by $\phi_{i}: G \twoheadrightarrow G_i$ the continuous canonical projection of the inverse limit described in Lemma~\ref{limitesdeS}. By assumption $G_i$ is generated by all its non-normal subgroups. Let $\bar{L}$ be one of these subgroups and consider $L_{i}$ its inverse image by $\phi_{i}$, clearly $L_{i}$ is a non-normal closed subgroup of $G$ containing $\ker(\phi_{i})$. Therefore $\ker(\phi_{i}) \leqslant \tilde{S}(G)$ and by Remark~\ref{quocientecappittprofinite} have $G/\ker(\phi_i) \neq \tilde{S}(G/\ker(\phi_i))=S(G/\ker(\phi_i)),$ so $|S(G_i)| < |G_i|<\infty$, a contradiction.
\end{proof}


In accordance with the notations of the previous lemmas we obtain the following.

\begin{lemm}
Let $G$ be a pro-$p$ Cappitt group. Then $G\,'$ is a procyclic central subgroup.
\label{commutador}
\end{lemm}

\begin{proof}
Since Cappitt groups are nilpotent of class at most $2$ (see Corollary~1 in~\cite{Por19}) we have $G'=\overline{[G,G]} \leqslant Z(G).$ Consider $G = \varprojlim_{i \in \mathcal{I}} \{G_i, \phi_{ij}\}$ where $\{G_i, \phi_{ij}, \mathcal{I}\}$ is a surjective inverse system of finite $p$-groups (with discrete topology). From Lemma~\ref{commutador2} it follows that $G_i \neq S(G_i)$, $\forall i \in \mathcal{I}$. Now, if any $G_i$ is abelian we have $(G_i)\,' =\{ 1\}$. On the other side, if $G_i$ is a non-abelian $p$-group then from Theorem~$1$ in~\cite{Cap71}, we have $(G_i)'$ is a finite cyclic $p$-group. Note that $\phi_i(G\,') = (G_i)\,', \forall i \in \mathcal{I}$, hence from Corollary~$1.1.8$ in~\cite{RZ10} we have $G\,' = \varprojlim_{i \in \mathcal{I}}\{(G_i)', \phi_{ij}\}$ whence it follows that $G\,'$ is procyclic.
\end{proof}

The preceding result is a generalization of the Corollary on page 314 in Cappitt~\cite{Cap71}. Kappe and Reboli show that the proof given by Cappitt was incorrect and give a new proof for this fact, this is the content of Theorem~4.2 in Kappe and Reboli~\cite{kappe}.


\begin{rema}\label{problemmm}
If $G$ is a non-abelian pro-$p$ Cappitt group, then $G$ is nilpotent of class $2$ (see Corollary~1 in~\cite{Por19}). In this case $G\,' \leqslant Z(G)$ and therefore $[[x,y],z]=1, \, \forall x, y, z \in G$. By Lemma~5.42$\MK$(i) in Rotman (see~\cite{rotman}), we have that $[x^n,y]=[x,y^n]=[x,y]^n, \, \forall n \in \mathbb{Z}.$
\end{rema}

\begin{lemm}\label{ecentral}
Let $G$ be a non-abelian pro-$p$ Cappitt group. If $a \in G \setminus \tilde{S}(G)$ has infinite order, then $a$ is central.
\end{lemm}
\begin{proof}
Let $a$ be an element in $G \setminus \tilde{S}(G)$ of infinite order, hence $|a|=p^{\infty}$ and $\overline{\left\langle a \right\rangle} \cong \mathbb{Z}_{p}$. If $a$ is not central in $G$ then there is $g \in G$ such that $[a,g] \neq 1$, moreover $g^{-1}ag \in \overline{\left\langle a \right\rangle}$ because $\overline{\left\langle a \right\rangle} \triangleleft G$. By Remark~\ref{problemmm} we have $g^{-1}aga^{-1}=[g,a^{-1}]={[g,a]}^{-1}=[a,g] \in \overline{\left\langle a \right\rangle}$. Since $\mathbb{Z}_p$ is a free pro-$p$ group on $\{1\}$, there is a unique continuous epimorphism $\zeta: \mathbb{Z}_p \to \overline{\left\langle a \right\rangle}$ such that $\zeta(1)=a$, therefore each element of $\overline{\left\langle a \right\rangle}$ can be written as $a^{\lambda} = \zeta(\lambda)$ for some $\lambda \in \mathbb{Z}_p$ (see similar notation in Section~4.1 on~\cite{RZ10}). Since $\overline{\left\langle a \right\rangle}$ is infinite it follows that $\zeta$ is an isomorphism (see Proposition~2.7.1 in~\cite{RZ10}). In particular $1 \neq [a,g]=a^{\delta}$ for some $\delta \in \mathbb{Z}_p$.
Let $(n_i)_{_{i \in \mathbb{N}}}$ be a sequence of integers converging to $\delta$ on $\mathbb{Z}_{p},$ in other words choose the $n_i$ such that $\lim_{i \to \infty } \, n_i= \delta$ (note that $ \overline{\left\langle a \right\rangle} = \mathbb{Z}_{p}$ is a complete metric space and $\overline{\mathbb{Z}} = \mathbb{Z}_{p}$). Adapting the argument from Lemma~4.1.1 in~\cite{RZ10} to $\mathbb{Z}_p$ we have $\lim_{i \to \infty } \,[a,g]^{n_i}= [a,g]^{\delta}.$ By properties of limits of sequences in metric spaces together with Remark~\ref{problemmm} we obtain
\[
[a,g]^{\delta}=\lim_{i \to \infty } \,[a^{n_i},g]=\lim_{i \to \infty }(a^{n_i})^{-1} \cdot \lim_{i \to \infty } g^{-1} \cdot \lim_{i \to \infty } a^{n_i} \cdot \lim_{i \to \infty } g=[a^{\delta},g].
\]
As $G'$ is central (see Remark~\ref{problemmm}) we have $[a,g]=a^{\delta} \in Z(G).$ So $[[a,g],g]=[a^{\delta},g]=[a,g]^{\delta}=1,$ which implies that $a^{2\cdot \delta }=\zeta(2 \cdot \delta)=1,$ a contradiction because $\Ker(\zeta)=\{0\}$. Therefore $a$ is central in $G.$
\end{proof}

\begin{prop}
Let $G$ be a non-abelian pro-$p$ Cappitt group and $\tor(G)$ is a closed subgroup of $G$. Then $G$ is a periodic group.
\label{torsionprop}
\end{prop}

\begin{proof}
Suppose that $G$ contains elements of infinite order. Since $G$ is nilpotent whose class is two (see Corollary~$1$ in~\cite{Por19}), it is generated by these elements, not all of which lie in $\tilde{S}(G)$ otherwise we would have $G= \overline{\left\langle G \setminus \tilde{S}(G) \right\rangle} =\overline{\left\langle \tor(G)\right\rangle}=\tor(G)$, a contradiction. Let $a$ be an element in $G \setminus \tilde{S}(G)$ of infinite order. From Lemma~\ref{ecentral} $a$ is central. Now if $w$ is a non-central element of $G \setminus \tilde{S}(G)$ then $aw, a^2w$ are non-central elements of infinite order, so both lie in $\tilde{S}(G)$. Thus $a^2ww^{-1}a^{-1}=a \in\tilde{S}(G)$, contradicting the choice of $a$. Therefore every element of $G \setminus \tilde{S}(G)$ is central and thus $G$ is abelian since $G \setminus \tilde{S}(G)$ generates $G$, a contradiction.
\end{proof}

\begin{exam*}
Note that the condition $\tilde{S}(G) \neq G$ in the Proposition~\ref{torsionprop} is necessary, for instance $G= \left(\prod_{r \in \mathbb{N}}\,Z_r\right) \times Q_8$ is a non-abelian pro-$2$ non-periodic group with $\tor(G) = Q_8$ and $Z_r \cong \mathbb{Z}_2$, $\forall r \in \mathbb{N}$. We show that $\tilde{S}(G)=G$. Fix $n \in \mathbb{N}$, indeed, it is sufficient we consider the following families of procyclic subgroups of $G$ isomorphic to $\mathbb{Z}_2$, that we will denote by
\begin{align*}
I_n&=\overline{\langle(0,\ldots, 0, 1_n, 0, \ldots,0, i)\rangle},
\\
J_n&=\overline{\langle(0,\ldots,0,1_n,0, \ldots,0,j)\rangle},
\\
K_n&=\overline{\langle(0,\ldots,0,1_n,0, \ldots,0,k)\rangle},
\end{align*}
where $1_n:=(0,\ldots,0,1_n,0,\ldots,1)$ is the generator of $Z_n \cong \mathbb{Z}_2$. Observe that
\[
(0,\ldots,0,1_n,0,\ldots,0,i)^j = (0,\ldots,0,1_n,0,\ldots,0,-i) \not \in I_n,
\]
so inductively $I_n$ is not a normal subgroup of $G$. Similarly, it can be shown also that $J_n$ and $K_n$ are not normal subgroups of $G$ for each $n \in \mathbb{N}$. Therefore by the definition of $\tilde{S}(G)$ we have that $I_n, \ J_n$ and $K_n$ are contained in $\tilde{S}(G)$. Now consider the following elements of $\tilde{S}(G)$, $x_n=(0,\ldots,0,1_n,0,\ldots, i)$ and $y_n=(0,\ldots,0,1_n,0,\ldots,0,j)$. Note that $x_n \cdot y_n^{-1} = (0,\ldots,0,-k):=-k$, so $k \in \tilde{S}(G)$. Similarly $i, j \in \tilde{S}(G)$ and this implies that $Q_8$ is contained in $\tilde{S}(G)$. Since $(0,\ldots,0,1_n,0,\ldots,0,k)$ and $(0, \ldots,0,-k)$ are in $\tilde{S}(G)$ we have $1_n:=(0, \ldots, 1_n,0,\ldots,0,1)$ belongs to $\tilde{S}(G)$, and therefore $\tilde{S}(G) = G$.
\label{exampleperiodic}
\end{exam*}


\section{Proof of the Theorem~\ref{teoA}}

We are now ready to prove a main result of this paper.

\begin{proof}
The fact that $G'$ is a procyclic central subgroup follows from Lemma~\ref{commutador}. If $G$ is Dedekind (i.e. pro-$2$ Hamiltonian group) the result follows immediately from Theorem~\ref{dedekindBP} and Remark~\ref{matalogo}. From now on we will consider that $\tilde{S}(G)\neq \{1\}$. Since $\tor(G)$ is closed in $G$, from Proposition~\ref{torsionprop} follows that $G$ is a pro-$p$ torsion group. If $G$ is finitely generated as pro-$p$ group then from celebrated Zel$'$manov Theorem (see Theorem~4.8.5c in~\cite{RZ10}) we have that $G$ is a finite $p$-group and the result follows. Therefore we can suppose that $G$ is an infinitely generated pro-$p$ and $\{1\} \neq \Tilde{S}(G) \neq G$. Also since $G$ is a torsion group and $G'$ is procyclic, it follows from Proposition~2.7.1~\cite{RZ10} that $G\,' = C_{p^n}$ for some $n \in \mathbb{N}$ fixed.

Suppose by contradiction that $G$ has no finite exponent. So there are $p$-elements of $G$, say, $x_1, x_ 2, \ldots, x_k, \ldots, (k \in \mathbb{N})$, such that $|x_k| \to \infty$ for $k \to \infty$ (elements of unlimited order).

Let $G^{ab}=G/\overline{[G,G]}$ be the abelianization of $G$. It is straightforward prove that $G^{ab}$ is an abelian pro-$p$ torsion group because $G$ is a torsion group. From Theorem~$4.3.8$ in~\cite{RZ10} we obtain that
\[
G^{ab} = \prod_{m(1)} C_p \times \prod_{m(2)} C_{p^2} \times \ldots \times \prod_{m(e)} C_{p^e},
\]
where each $m(i)$ is a cardinal number, $e$ is some natural number, with at least one $m(i)$ infinite because $G^{ab}$ is infinitely generated.

For each $g \in G$ we have $(gG\,')^{p^e} = G\,'$ since $p^e$ is the finite exponent of $G^{ab}$. Therefore for all $k \in \mathbb{N}$ we have
\[
G\,' = (x_kG\,')^{p^e}=x_k^{p^e}G' \Longrightarrow x_k^{p^e} \in G\,'=C_{p^n}.
\]
Thus we obtain
\[
({x_k}^{p^e})^{p^n}= x_k^{p^{e+n}}=1.
\]
Therefore the order of each $x_k$ is limited by $p^{e+n}$, $\forall k \in \mathbb{N},$ a contradiction. We conclude that $G$ has finite exponent.
\end{proof}


\section{Pro-\texorpdfstring{$2$}{2} Cappitt groups of exponent \texorpdfstring{$4$}{4}}

The following result shows that the only pro-$2$ Cappitt groups of exponent $4$ are pro-$2$ Dedekind groups.

\begin{ltheo}\label{teoB}
If $G$ is a pro-$2$ group of exponent $4$ then either $\tilde{S}(G)=\{1\}$ or $\tilde{S}(G)=G$.
\end{ltheo}

\begin{proof}
Suppose that $\{1\} \neq \tilde{S}(G) \neq G$. It is clear that not every procyclic subgroup of $G$ is normal, otherwise $\tilde{S}(G) = \{1\}$ (see Remark~\ref{remark prociclica}). Then consider $J = \overline{\langle x \rangle}$ a non-normal procyclic subgroup of $G$. Let $H= \overline{\langle x,y,z\rangle} \leqslant_{c} G$ such that $y \not\in N_{G}(J)$ and $z \not\in \tilde{S}(G)$. By construction and Remark~\ref{quocientecappitt} we have that $H$ is a $3$-generated group of exponent $4$ with $\{1\} \neq \tilde{S}(H) \neq H$ and $\overline{\langle z\rangle} \trianglelefteq_{c} H$. Since $H$ has exponent $4$, we have that $H$ is a finitely generated pro-$2$ torsion group, it follows from celebrated Zel$'$manov Theorem that $H$ is a finite non-abelian $2$-group (see Theorem~4.8.5c in~\cite{RZ10}). From Corollary~1 in Porto~\cite{Por19} and Lemma~\ref{commutador} we have that $G$ and $H$ are nilpotent groups of class $2$ with $H'$ being cyclic, so $H' \cong C_2$ or $H' \cong C_4$. Take a minimal set of generators of $H$ chosen to lie outside $\tilde{S}(H)=S(H)$, say, $\{h_1, h_2, \ldots, h_l\}$, where $l \geq 2$. Note that $H'=\left \langle [h_i, h_j]\,\mid \,i,j \in\{1,2, \ldots,l\}, i \neq j\right \rangle$ (see p.~129 in~\cite{robinson}). Let $H_i=\left \langle h_i \right \rangle$ for each $i \in \{1,2, \ldots,l\}$. Since $h_i \not \in S(H)$ we have $H_i \trianglelefteq H$ and so $[h_i, h_j] \in H_i \cap H_j$ for each $i,j \in\{1,2, \ldots,l\}$.
Note that $[h_i, h_j]$ belongs to $H_m$ for every $i, j, m\in \{1,2,\ldots,l\}$, so $H'\leqslant H_m$ for each $m\in \{1,2,\ldots,l\}$. If $H' \cong C_4$ we have that $H'=H_1=H_2=\cdots=H_m$, a contradiction since $H$ is not abelian. Therefore $H'$ has order $2$. Note that $H$ is a finite $2$-group and it is $3$-generated with exponent $4$, so it is well-know that its order is at most $64$. A GAP computation using~\cite{Gap} shows that $H$ is either Dedekind or $\tilde{S}(H)=S(H) = H$, a contradiction.
\end{proof}

For example, a GAP~\cite{Gap} check yields, the SmallGroup(16, 6), SmallGroup(27, 4) and SmallGroup(64, 28) are finite Cappitt groups with commutator subgroup $C_2$, $C_3$ and $C_4$, respectively.

In Corollary~$4.4$ in~\cite{kappe}, Kappe and Reboli prove that if $G$ is a $p$-group Cappitt, then $G'$ is a finite cyclic $p$-group. Finally, in view of this result, example above, Lemma~\ref{commutador} and Theorem~\ref{teoA}, we can pose the following questions.

\section{Open questions about profinite Cappitt groups}
\begin{enumerate}
\item Are there non-abelian non-periodic pro-$p$ Cappitt groups? If $G$ is not torsion, is $G$ virtually $p$-adic?

\item If $G$ is a pro-$p$ group satisfying $\{1\}\neq \tilde{S}(G) \neq G$, do we always have that $G' \cong C_{p^n}$ for some $n \in \mathbb{N}$?
\end{enumerate}

\subsection*{Acknowledgment} We thank John MacQuarrie for the careful reading and suggestions.



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